Let \( P(\alpha, \beta) \) be a point on the parabola \( y^2 = 4x \). If \( P \) also lies on the chord of the parabola \( x^2 = 8y \) whose midpoint is \( \left( 1, \frac{5}{4} \right) \), then \( (\alpha - 28)(\beta - 8) \) is equal to ______.
Correct Answer: 192
Approach Solution - 1
Step 1: Parabola equation and midpoint of the chord.
The given parabola is \( x^2 = 8y \), and the midpoint \( (x_1, y_1) = (1, \frac{5}{4}) \) lies on the chord. The equation of the chord is given by: \[ x x_1 + 4(y + y_1) = x_1^2 - 8y_1 \] Substituting the values \( x_1 = 1 \) and \( y_1 = \frac{5}{4} \) into the equation: \[ x \cdot 1 + 4(y + \frac{5}{4}) = 1^2 - 8 \cdot \frac{5}{4} \] Simplifying: \[ x - 4y + 4 = -9 \quad \Rightarrow \quad x - 4y = -13 \] This is our equation (i): \[ x - 4y + 4 = 0 \quad \dots \text{(i)} \]
Step 2: Substituting the point \( (\alpha, \beta) \) into the equation.
The point \( (\alpha, \beta) \) lies on both the line \( x - 4y + 4 = 0 \) and the parabola \( y^2 = 4x \), so we substitute \( \alpha \) and \( \beta \) into these equations. From equation (i), we get: \[ \alpha - 4\beta + 4 = 0 \quad \Rightarrow \quad \alpha = 4\beta - 4 \quad \dots \text{(ii)} \] From the equation of the parabola \( y^2 = 4x \), we have: \[ \beta^2 = 4\alpha \quad \dots \text{(iii)} \]
Step 3: Solving the system of equations.
Substitute equation (ii) into equation (iii): \[ \beta^2 = 4(4\beta - 4) \] \[ \beta^2 = 16\beta - 16 \] \[ \beta^2 - 16\beta + 16 = 0 \] Solving this quadratic equation for \( \beta \): \[ \beta = \frac{16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{16 \pm \sqrt{256 - 64}}{2} = \frac{16 \pm \sqrt{192}}{2} = \frac{16 \pm 8\sqrt{3}}{2} \] \[ \beta = 8 \pm 4\sqrt{3} \] Thus, the two possible values for \( \beta \) are: \[ \beta = 8 + 4\sqrt{3} \quad \text{or} \quad \beta = 8 - 4\sqrt{3} \]
Step 4: Solving for \( \alpha \).
Substitute \( \beta = 8 + 4\sqrt{3} \) into equation (ii) to find \( \alpha \): \[ \alpha = 4(8 + 4\sqrt{3}) - 4 = 28 + 16\sqrt{3} \] So, \( \alpha = 28 + 16\sqrt{3} \).
Step 5: Calculating \( (\alpha - 28)(\beta - 8) \).
Now, we calculate \( (\alpha - 28)(\beta - 8) \): \[ (\alpha - 28)(\beta - 8) = (16\sqrt{3})(4\sqrt{3}) = 16 \times 4 \times 3 = 192 \]
Final Answer:
\[ \boxed{192} \]
Approach Solution -2
Step 1. The equation of the parabola is \(x^2 = 8y\)
Step 2. The chord with midpoint \( (x_1, y_1) \) has the equation \( T = S_1 \):
\(x x_1 - 4(y + y_1) = x_1^2 - 8y_1\)
Substituting \( (x_1, y_1) = (1, \frac{5}{4}) \):
\(x - 4 \left( y + \frac{5}{4} \right) = 1 - 8 \cdot \frac{5}{4} = -9\)
\(x - 4y = -4\)
Step 3. Since \( P(\alpha, \beta) \) lies on this chord and also on the parabola \( y = \frac{x^2}{4} \), we have:
\(\alpha - 4\beta = -4\)
\(\beta^2 = 4\alpha\)
Step 4. Solve equations (ii) and (iii):
Substitute \( \alpha = \frac{\beta^2}{4} \) from (iii) into (ii):
\(\frac{\beta^2}{4} - 4\beta = -4\)
\(\beta^2 - 16\beta + 16 = 0\)
\((\beta - 8)^2 = 48\)
\(\beta = 8 \pm 4\sqrt{3}\)
\(\beta = 8 \pm 4\sqrt{3}\)
Step 5. Substitute \( \beta = 8 \pm 4\sqrt{3} \) back into equation (ii) to find \( \alpha \):
For \(\beta = 8 + 4\sqrt{3}\):
For \(\beta = 8 - 4\sqrt{3}\):
\(\alpha = 4(8 - 4\sqrt{3}) - 4 = 28 - 16\sqrt{3}\)
Step 6. Therefore, the possible points \( ( \alpha, \beta ) \) are:
\(( \alpha, \beta ) = (28 + 16\sqrt{3}, 8 + 4\sqrt{3}) \text{ and } (28 - 16\sqrt{3}, 8 - 4\sqrt{3})\)
Step 7. Calculate \( ( \alpha - 28 )( \beta - 8 ) \):
\(( \alpha - 28 )( \beta - 8 ) = (\pm 16\sqrt{3})(\pm 4\sqrt{3}) = 16 \cdot 4 \cdot 3 = 192\)
Thus, \(( \alpha - 28 )( \beta - 8 ) = 192\).
The Correct Answer is: 192
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